R E S E A R C H
Open Access
On the existence of fixed points that
belong to the zero set of a certain function
Erdal Karapinar
1,2, Donal O’Regan
2,3and Bessem Samet
4**Correspondence: [email protected] 4Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh, 11451, Saudi Arabia
Full list of author information is available at the end of the article
Abstract
LetT:X→Xbe a given operator andFTbe the set of its fixed points. For a certain function
ϕ
:X→[0,∞), we say thatFTisϕ-admissible if
FTis nonempty andFT⊆Zϕ, whereZϕis the zero set ofϕ. In this paper, we study the
ϕ-admissibility of a new class
of operators. As applications, we establish a new homotopy result and we obtain a partial metric version of the Boyd-Wong fixed point theorem.MSC: 54H25; 47H10
Keywords:
ϕ-admissible; fixed point; homotopy result; partial metric
1 Introduction
Let (X,d) be a metric space. For a given functionϕ:X→[,∞), we define the set
Zϕ=
x∈X:ϕ(x) = .
LetT :X→Xbe a given operator. The set of fixed points ofT is denoted byFT, that
is,
FT={x∈X:Tx=x}.
Definition . We say that the setFTisϕ-admissible if and only ifFT=∅andFT⊆Zϕ.
Let F be the set of functions F : [,∞) →[,∞) satisfying the following
condi-tions:
(F) max{a,b} ≤F(a,b,c), for alla,b,c≥; (F) F(a, , ) =a, for alla≥;
(F) Fis continuous.
As examples, the following functions belong toF:
. F(a,b,c) =a+b+c,
. F(a,b,c) =max{a,b}+ln(c+ ), . F(a,b,c) =a+b+c(c+ ), . F(a,b,c) = (a+b)ec,
. F(a,b,c) = (a+b)(c+ )n,n∈N.
Letbe the set of functionsψ: [,∞)→[,∞) satisfying the following conditions:
() ψ is upper semi-continuous from the right; () ψ(t) <t, for allt> .
For given functionsϕ:X→[,∞),F∈F, andψ∈, we denote byT(ϕ,F,ψ) the class of operatorsT:X→Xsatisfying
Fd(Tx,Ty),ϕ(Tx),ϕ(Ty)≤ψFd(x,y),ϕ(x),ϕ(y), (x,y)∈X×X. (.)
The aim of this paper is to study theϕ-admissibility of the setFT, whereTbelongs to
the class of operatorsT(ϕ,F,ψ), (F,ψ)∈F×. As applications, we obtain an homotopy result and a partial metric version of the Boyd-Wong fixed point theorem.
2 Main result
Our main result is given in the following theorem.
Theorem . Let(X,d)be a complete metric space and T:X→X be a given operator.
Suppose that the following conditions hold:
(i) there existϕ:X→[,∞),F∈F,andψ∈such thatT∈T(ϕ,F,ψ); (ii) ϕis lower semi-continuous.
Then the set FTisϕ-admissible.Moreover,the operator T has a unique fixed point.
Proof Letξ be an arbitrary element of the setFT. Takex=y=ξin (.), and we get
F,ϕ(ξ),ϕ(ξ)≤ψF,ϕ(ξ),ϕ(ξ). (.)
IfF(,ϕ(ξ),ϕ(ξ))= , from (), we get
ψF,ϕ(ξ),ϕ(ξ)<F,ϕ(ξ),ϕ(ξ),
which is impossible from (.). Consequently, we have
F,ϕ(ξ),ϕ(ξ)= .
Using the above equality and (F), we obtain
ϕ(ξ)≤F,ϕ(ξ),ϕ(ξ)= ,
which yields
ϕ(ξ) = .
Consequently, we have
Now, we have to prove thatFT is a nonempty set. Letxbe an arbitrary element ofX.
Consider the Picard sequence{xn} ⊂Xdefined by
xn=Tnx, n∈N={, , , . . .},
whereTnis thenth iterate ofT. If for someN∈Nwe havexN=xN+, thenxN will be an
element ofFT. As a result we can suppose that
d(xn,xn+) > , n∈N. (.)
Using (.), we have
Fd(Txn,Txn–),ϕ(Txn),ϕ(Txn–)
≤ψFd(xn,xn–),ϕ(xn),ϕ(xn–)
, n∈N∗; (.)
hereN∗={, , . . .}. If for someN∈N∗, we have
Fd(xN,xN–),ϕ(xN),ϕ(xN–)
= ,
then property (F) yields
d(xN,xN–)≤F
d(xN,xN–),ϕ(xN),ϕ(xN–)
= ,
which is a contradiction with (.). Thus
Fd(xn,xn–),ϕ(xn),ϕ(xn–)
> , n∈N∗. (.)
Using (.), (.), the definition of the sequence{xn}, and (), we have
⎧ ⎨ ⎩
F(d(xn+,xn),ϕ(xn+),ϕ(xn))≤ψ(F(d(xn,xn–),ϕ(xn),ϕ(xn–))),
ψ(F(d(xn,xn–),ϕ(xn),ϕ(xn–))) <F(d(xn,xn–),ϕ(xn),ϕ(xn–)),
n∈N∗. (.)
It follows immediately from (.) that there exists somec≥ such that
lim
n→∞F
d(xn+,xn),ϕ(xn+),ϕ(xn)
= lim
n→∞ψ
Fd(xn,xn–),ϕ(xn),ϕ(xn–)
=c. (.)
Suppose now thatc> . Using the properties ()-(), we deduce from (.) that
c=lim sup
n→∞
ψFd(xn,xn–),ϕ(xn),ϕ(xn–)
≤ψ(c) <c,
which is a contradiction. As a consequence, we havec= , that is,
lim
n→∞F
d(xn+,xn),ϕ(xn+),ϕ(xn)
= lim
n→∞ψ
Fd(xn,xn–),ϕ(xn),ϕ(xn–)
Using (F) and (.), we get
lim
n→∞d(xn+,xn) =nlim→∞ϕ(xn) = . (.)
Next, we show that{xn}is a Cauchy sequence in the metric space (X,d). Suppose that{xn}
is not a Cauchy sequence. Then there existsε> for which we can find two sequences of positive integers{m(k)}and{n(k)}such that, for allk∈N,
n(k) >m(k) >k, d(xm(k),xn(k))≥ε, d(xm(k),xn(k)–) <ε. (.)
Using (.), for allk∈Nwe have
ε≤d(xm(k),xn(k))
≤d(xm(k),xn(k)–) +d(xn(k)–,xn(k))
< ε+d(xn(k)–,xn(k)),
which yields
ε≤d(xm(k),xn(k)) <ε+d(xn(k)–,xn(k)), k∈N. (.)
Lettingk→ ∞in the above inequality and using (.), we obtain
lim
k→∞d(xm(k),xn(k)) =ε
+ i.e. lim
k→∞d(xm(k),xn(k)) =ε and d(xm(k),xn(k))≥ fork∈N.
(.)
Using the properties (F)-(F), (.), and (.), we get
lim
k→∞F
d(xn(k),xm(k)),ϕ(xn(k)),ϕ(xm(k))
=F(ε, , ) =ε+.
Using the above limit and (), we obtain
lim sup
k→∞ ψ
Fd(xn(k),xm(k)),ϕ(xn(k)),ϕ(xm(k))
≤ψ(ε). (.)
On the other hand, using (.) and (F), for allk∈Nwe have
ε≤d(xn(k),xm(k))≤d(xn(k),xn(k)+) +d(xn(k)+,xm(k)+) +d(xm(k)+,xm(k))
≤d(xn(k),xn(k)+) +F
d(xn(k)+,xm(k)+),ϕ(xn(k)+),ϕ(xm(k)+)
+d(xm(k)+,xm(k))
≤d(xn(k),xn(k)+) +ψ
Fd(xn(k),xm(k)),ϕ(xn(k)),ϕ(xm(k))
+d(xm(k)+,xm(k)).
Passing to the limit superior ask→ ∞, using (.), (.), and (), we obtain
which is a contradiction. As a consequence,{xn}is a Cauchy sequence. Since (X,d) is a
complete metric space, there is az∈Xsuch that
lim
n→∞d(xn,z) = . (.)
Sinceϕis lower semi-continuous, it follows from (.) and (.) that
≤ϕ(z)≤lim inf
n→∞ ϕ(xn) = ,
which yields
z∈Zϕ. (.)
Now we show thatz∈FT. Using (.), (F), and (.), we have
d(xn+,Tz)≤ψ
Fd(xn,z),ϕ(xn),
, n∈N. (.)
Also using the continuity ofF, (F), (.), and (.), we have
lim
n→∞F
d(xn,z),ϕ(xn),
=F(, , ) = .
Note that from (), we have
lim
t→+ψ(t) = . Then
lim
n→∞ψ
Fd(xn,z),ϕ(xn),
= lim
t→+ψ(t) = . (.)
Now, passingn→ ∞in (.) and using (.), we get
lim
n→∞d(xn+,Tz) = .
The uniqueness of the limit yields z =Tz. Thus FT is a nonempty set, and the ϕ
-admissibility ofFTis proved. Finally, in order to prove the uniqueness of the fixed point, let
us assume thatw∈FTwithd(z,w) > . SinceFTisϕ-admissible, we know thatz,w∈Zϕ.
Now, applying (.) with (x,y) = (z,w), we obtain
Fd(z,w), , ≤ψFd(z,w), , .
Using the properties (F) and (), we get
d(z,w)≤ψd(z,w)<d(z,w),
Remark . Takeϕ≡ andF(a,b,c) =a+b+cin Theorem ., and we recover the Boyd-Wong fixed point theorem [].
Now, we give some examples to illustrate our main result given by Theorem ..
Example . We endow the setX= [,∞) with the standard metric
d(x,y) =|x–y|, (x,y)∈X×X. LetT:X→Xbe the mapping defined by
Tx=
⎧ ⎨ ⎩
if ≤x≤,
x
if <x.
Observe thatTis not continuous inX. So, there is noψ∈such that
d(Tx,Ty)≤ψd(x,y), (x,y)∈X×X.
Then the Boyd-Wong fixed point theorem cannot be applied in this case. Let ϕ:X→
[,∞) be the function defined by
ϕ(x) =xn, x∈X, for somen∈N∗.
LetF: [,∞)→[,∞) be the function defined by
F(a,b,c) =a+b+c, a,b,c≥.
Letψ: [,∞)→[,∞) be the function defined by
ψ(t) = t
, t≥.
Observe that F belongs to the set F andψ belongs to the set . We claim that T ∈
T(ϕ,F,ψ), that is,
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty)≤
d(x,y) +ϕ(x) +ϕ(y)
, (x,y)∈X×X. (.) In order to prove our claim, we distinguish three cases.
Case. (x,y)∈[, ]×[, ]. In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) = ≤
d(x,y) +ϕ(x) +ϕ(y)
.
Case. (x,y)∈[, ]×(,∞). In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) =y +
y
n
while
d(x,y) +ϕ(x) +ϕ(y)
=
y–x+x
n+yn.
Then we have to prove that
yn
n––
≤xn–x.
Observe that the functionh: [, ]→Rdefined by
h(x) =xn–x, x∈[, ],
has a global minimum atxn= (n)
n– which is equal toxn(–n
n )≥
–n
n . So, we have just to
check that
yn
n––
≤
n– .
Sincey> , we have
yn
n––
≤
n–– ≤
n– .
Then our claim holds in this case.
Case.x,y> . In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) =|x–y|
+
x
n
+
y
n
and
d(x,y) +ϕ(x) +ϕ(y)
=|x–y|
+
xn+yn
.
Obviously, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty)≤
d(x,y) +ϕ(x) +ϕ(y)
.
Finally, in all cases our claim (.) holds, which yieldsT∈T(ϕ,F,ψ). By Theorem ., the setFT isϕ-admissible andT has a unique fixed point. In this example,FT={}and
ϕ() = .
Example . We endow the setX= [√,∞) with the standard metric
LetT:X→Xbe the mapping defined by
Tx=
⎧ ⎨ ⎩
√
if√≤x≤√,
x
if
√ <x.
As in the previous example, the Boyd-Wong fixed point theorem cannot be applied in this case. Letϕ:X→[,∞) be the function defined by
ϕ(x) =x– , x∈X.
LetF: [,∞)→[,∞) be the function defined by
F(a,b,c) =a+b+c, a,b,c≥.
Letψ: [,∞)→[,∞) be the function defined by
ψ(t) = t
, t≥. We distinguish three cases.
Case. (x,y)∈[√, √]×[√, √]. In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) = ≤
d(x,y) +ϕ(x) +ϕ(y)
.
Case. (x,y)∈[√, √]×(√,∞). In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) =y –
√ +y
– ,
while
d(x,y) +ϕ(x) +ϕ(y)
=y –
x
+
x
+
y
– .
Clearly, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty)≤
d(x,y) +ϕ(x) +ϕ(y)
.
Case. (x,y)∈(√,∞). In this case, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty) =|x–y|
+
x
+
y
– ,
while
d(x,y) +ϕ(x) +ϕ(y)
=|x–y|
+
x
+
y
Also, we have
d(Tx,Ty) +ϕ(Tx) +ϕ(Ty)≤
d(x,y) +ϕ(x) +ϕ(y)
.
As a consequence, the mappingT belongs toT(ϕ,F,ψ). By Theorem ., the setFT is
ϕ-admissible andThas a unique fixed point. In this example,FT={
√
}andϕ(√) = .
Example . Let (X,d) be the metric space considered in Example .. We take the func-tionsϕ,F, andψ defined in Example .. LetT:X→Xbe the mapping defined by
Tx=
⎧ ⎨ ⎩
√
if√≤x≤√, sinx
if
√ <x.
Similarly, we haveT∈T(ϕ,F,ψ). By Theorem ., the setFTisϕ-admissible andThas a
unique fixed point. In this example,FT={
√
}andϕ(√) = .
3 Applications
3.1 An homotopy result
Let us denote byF∗the set of functionsF∈Fsatisfying the following property:
(F) for alla,b,c,d≥,
a≤c+d ⇒ F(a,b, )≤F(c,b, ) +d. As examples, the following functions belong toF∗:
. F(a,b,c) = (a+b)ec,
. F(a,b,c) = (a+b)(c+ )n,n∈N.
Observe thatF∗F. To see this, let us consider the function
F(a,b,c) =aec+b+bea+c, a,b,c≥. It is not difficult to check thatF∈FbutF∈/F∗.
We have the following homotopy result.
Theorem . Let(X,d)be a complete metric space, U be an open subset of X,and V be a closed subset of X with U ⊂V.Suppose that H:V ×[, ]→X has the following properties:
(C) x=H(x,λ)for everyx∈V\Uandλ∈[, ];
(C) there exist a continuous functionϕ:X→[,∞),L∈(, ),andF∈F∗such that for allx,y∈Vandλ∈[, ],
FdH(x,λ),H(y,λ),ϕH(x,λ),ϕH(y,λ)≤LFd(x,y),ϕ(x),ϕ(y);
(C) there exists a continuous functionη: [, ]→Rsuch that for allx∈Vandλ,μ∈[, ],
FdH(x,λ),H(x,μ),ϕH(x,λ),ϕH(x,μ)≤η(λ) –η(μ).
Proof Suppose thatH(·, ) has a fixed point. Consider the set
Q=t∈[, ] :x=H(x,t) for somex∈U.
From (C), clearly is an element of Q, soQis a nonempty set. We will show that Q
is both closed and open in [, ], and so by the connectedness of [, ], we are finished sinceQ= [, ]. First, let us prove that Qis open in [, ]. Lett∈Qandx∈U with
x=H(x,t). Using (C) withx=y=xandλ=t, we obtain
F,ϕ(x),ϕ(x)
≤LF,ϕ(x),ϕ(x)
,
which implies sinceL∈(, ) that
F,ϕ(x),ϕ(x)
= .
Then (F) yields
ϕ(x) = .
Moreover, observe that, for allt∈[, ], ifx∈Uis a fixed point ofH(·,t), thenϕ(x) = . On the other hand, sinceUis open in (X,d), there existsr> such thatB(x,r)⊆U, where
B(x,r) =
z∈X:d(x,z) <r
.
Consider the set
(x,ϕ) =
z∈X:Fd(z,x),ϕ(z),
<r.
Clearly (x,ϕ) is nonempty (since x ∈ (x,ϕ)) and (x,ϕ)⊆ B(x,r). Let ε =
( –L)r> . Sinceηis continuous ont, there existsα(ε) > such that
t∈t–α(ε),t+α(ε)
∩[, ] ⇒ η(t) –η(t)<ε.
Lett∈(t–α(ε),t+α(ε))∩[, ]. Forx∈(x,ϕ) (the closure of(x,ϕ)), we have
FdH(x,t),x
,ϕH(x,t), =FdH(x,t),H(x,t)
,ϕH(x,t), . Also since
dH(x,t),H(x,t)
≤dH(x,t),H(x,t)
+dH(x,t),H(x,t)
,
using the properties (F), (F) we get
FdH(x,t),H(x,t)
,ϕH(x,t), ≤FdH(x,t),H(x,t)
,ϕH(x,t), +dH(x,t),H(x,t)
≤FdH(x,t),H(x,t)
,ϕH(x,t), +FdH(x,t),H(x,t)
,ϕH(x,t)
≤η(t) –η(t)+LF
d(x,x),ϕ(x),
<ε+Lr=r.
Thus we proved that, for allt∈(t–α(ε),t+α(ε))∩[, ], the operator
H(·,t) :(x,ϕ)→(x,ϕ)
is well defined. Now, using (C) and Theorem ., we deduce that, for allt∈(t–α(ε),
t+α(ε))∩[, ], the operatorH(·,t) has a fixed point inV. However, such a fixed point
should be inUfrom (C). As a consequence,
t–α(ε),t+α(ε)
∩[, ]⊆Q,
which proves thatQis open in [, ]. Next, we show thatQis closed in [, ]. To see this, let{tn}be a sequence inQwithtn→t∈[, ] asn→ ∞. We have to prove thatt∈Q.
From the definition ofQ, for alln∈N, there existsxn∈Uwith
xn=H(xn,tn) and ϕ(xn) = .
Also for allm,n∈N, we have
d(xn,xm) =d
H(xn,tn),H(xm,tm)
≤dH(xn,tn),H(xn,tm)
+dH(xn,tm),H(xm,tm)
≤FdH(xn,tn),H(xn,tm)
,ϕH(xn,tn)
,ϕH(xn,tm)
+FdH(xn,tm),H(xm,tm)
,ϕH(xn,tm)
,ϕH(xm,tm)
≤η(tn) –η(tm)+LF
d(xn,xm), ,
=η(tn) –η(tm)+Ld(xn,xm),
which yields
d(xn,xm)≤|
η(tn) –η(tm)|
–L , m,n∈N.
Lettingm,n→ ∞in the above inequality and using the continuity ofη, we getd(xn,xm)→
asm,n→ ∞, which implies that{xn}is a Cauchy sequence in the complete metric space
(X,d). Then there is somez∈V(sinceV is closed) such that
lim
n→∞d(xn,z) = and ϕ(z) = ,
sinceϕis lower semi-continuous. Now, for alln∈Nwe have
dxn,H(z,t)
=dH(xn,tn),H(z,t)
≤dH(xn,tn),H(xn,t)
+dH(xn,t),H(z,t)
≤FdH(xn,tn),H(xn,t)
,ϕH(xn,tn)
,ϕH(xn,t)
+FdH(xn,t),H(z,t)
,ϕH(xn,t)
≤η(tn) –η(t)+LF
d(xn,z), ,
=η(tn) –η(t)+Ld(xn,z).
Lettingn→ ∞in the above inequality, we obtain
lim
n→∞d
xn,H(z,t)
= .
The uniqueness of the limit yieldsz=H(z,t). Using (C), we deduce thatz∈Uandt∈Q. ThusQis closed in [, ].
For the reverse implication, we use the same technique.
3.2 A partial metric version of Boyd-Wong fixed point theorem
In this part, using Theorem ., we obtain a partial metric version of the Boyd-Wong fixed point theorem.
We start by recalling some basic definitions and properties of partial metric spaces. For more details of such spaces, we refer the reader to [–].
A partial metric on a nonempty setXis a functionp:X→X→[,∞) such that for all
x,y,z∈X, we have
(i) p(x,x) =p(y,y) =p(x,y)⇐⇒x=y; (ii) p(x,x)≤p(x,y);
(iii) p(x,y) =p(y,x);
(iv) p(x,y)≤p(x,z) +p(z,y) –p(z,z).
A partial metric space is a pair (X,p) such thatXis a nonempty set andpis a partial metric onX. It is clear that, ifp(x,y) = , then from (i)-(ii),x=y; but ifx=y,p(x,y) may not be . A basic example of a partial metric space is the pair ([,∞),p), wherep(x,y) =max{x,y}.
Each partial metric ponXgenerates a Ttopologyτp onX which has as a base the
family of openp-balls{Bp(x,ε) :x∈X,ε> }, where
Bp(x,ε) :=
y∈X:p(x,y) <p(x,x) +ε.
Let (X,p) be a partial metric space. A sequence{xn} ⊂Xconverges to somex∈Xwith
respect topif and only if
lim
n→∞p(xn,x) =p(x,x).
A sequence{xn} ⊂X is said to be a Cauchy sequence if and only iflimm,n→∞p(xn,xm)
exists and is finite. The partial metric space (X,p) is said to be complete if and only if every Cauchy sequence{xn}inXconverges to somex∈Xsuch thatlimn,m→∞p(xn,xm) =p(x,x).
Ifpis a partial metric onX, then the functiondp:X→X→[,∞) defined by
dp(x,y) = p(x,y) –p(x,x) –p(y,y), (x,y)∈X, (.)
is a metric onX.
Lemma . Let(X,p)be a partial metric space.Then:
(i) {xn}is a Cauchy sequence in(X,p)if and only if{xn}is a Cauchy sequence in the
(ii) the partial metric space(X,p)is complete if and only if the metric space(X,dp)is
complete.Furthermore,limn→∞dp(xn,x) = if and only if
lim
n→∞p(xn,x) =p(x,x) =m,limn→∞p(xn,xm).
We have the following result.
Corollary . Let(X,p)be a complete partial metric space and let T:X→X be an oper-ator such that
p(Tx,Ty)≤ψp(x,y), (x,y)∈X×X, (.)
whereψ∈.We have the following results:
(i) ifz∈Xis a fixed point ofTthenp(z,z) = ; (ii) T has a unique fixed point.
Proof Letdpbe the metric onXdefined by (.). We have
p(x,y) =d(x,y) +ϕ(x) +ϕ(y), (x,y)∈X×X,
where
d(x,y) = dp(x,y)
, ϕ(x) =
p(x,x) .
Then (.) yields
Fd(Tx,Ty),ϕ(Tx),ϕ(Ty)≤ψFd(x,y),ϕ(x),ϕ(y), (x,y)∈X×X,
where
F(a,b,c) =a+b+c, a,b,c≥.
From (ii) Lemma ., the metric space (X,d) is complete and the functionϕis continuous with respect to the topology ofd. Finally the desired result follows from Theorem ..
Remark . Take in Corollary .,ψ(t) =ktwith k∈(, ), and we recover Matthews fixed point theorem [].
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Atilim University, Incek, Ankara, 06836, Turkey.2Nonlinear Analysis and Applied
Acknowledgements
The third author would like to extend his sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding of this research through the International Research Group Project No. IRG14-04.
Received: 16 May 2015 Accepted: 10 August 2015
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